Question: Find $\lim_{h\to 0}\dfrac{9\cdot 2^{{1+h}}-9\cdot 2^1}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $9$ (Choice B) B $18$ (Choice C) C $18\ln(2)$ (Choice D) D The limit doesn't exist
Solution: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $9\cdot 2^{{1+h}}-9\cdot 2^1$, we can tell that the function is $f(x)=9\cdot 2^x$ and the $x$ -value is $1$. In other words, the limit expression is equal to $f'(1)$ for $f(x)=9\cdot 2^x$. Let's find $f'(x)$ : $f'(x)=9\cdot \ln(2)\cdot 2^x$ Now let's evaluate $f'(1)$ : $f'(1)=9\cdot \ln(2)\cdot 2^1=18\ln(2)$ In conclusion, $\lim_{h\to 0}\dfrac{9\cdot 2^{{1+h}}-9\cdot 2^1}{h}=18\ln(2)$.